Question

The distance moved by a freely falling body
(starting from rest) during the 1st, 2nd and 3rd
... nth second of its motion, are proportional to​

Answer 1

Answer:

The distance moved by a freely falling body

(starting from rest) during the 1st, 2nd and 3rd

... nth second of its motion, are proportional to time

Answer 2

Answer:

The distance moved by a freely falling body during the 1st, 2nd and 3rd second of its motion, are proportional to​ odd numbers.

Explanation:

Given, the freely falling body was initially at rest, u=0

We know,

Distance travelled by a body is given by 2nd equation of motion:

$S=u+\frac{1}{2}at^{2}$

Put u=0 and a=g in the above equation:

$S_{n}=\frac{g}{2}t^{2}$                                                    ..................(1)

Distance travelled in 1st second will be:

$S_{1}=\frac{1}{2}g(1)^{2}$

$S_{1}=\frac{1}{2}g$

Distance travelled in 2nd  second will be:

$S_{2}=\frac{1}{2}g(2)^{2}$

$S_{2}=\frac{4}{2}g$

Distance travelled in 3rd second will be:

$S_{3}=\frac{1}{2}g(3)^{2}$

$S_{3}=\frac{9}{2}g$

Therefore, Distance D₁ = S₁ $=\frac{g}{2}$

Distance D₂ = S₂-S₁ $=\frac{4}{2}g-\frac{1}{2} g$

$D_{2}=g(2 -\frac{1}{2} )$

$D_{2}=\frac{3}{2}g$

Distance D₃= S₃-S₂ $=\frac{9}{2}g-2g$

$D_{3}=g(\frac{9}{2} -2)$

$D_{3}=\frac{5}{2} g$

The ratio of D₁ : D₂ : D₃ will be:

⇒   $\frac{1}{2} g:\frac{3}{2} g:\frac{5}{2} g$

⇒   $1 : 3: 5$

Therefore the distance moved by body  D₁, D₂ and D₃ are proportional to odd numbers.